Question

An electric bulb is rated 200 W - 240 V. Find the resistance of the filament of the bulb​

Answer 1

Answer:

⠀⠀⠀⠀

⠀⠀⠀⠀

Resistance of the filament of the bulb is 288 Ω.

⠀⠀⠀⠀

⠀⠀⠀⠀

Explanation:

⠀⠀⠀⠀

⠀⠀⠀⠀

Given:

⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀

The Bulb is rated 200 W - 240 V

⠀⠀⠀⠀

• Power (P) of the elctric bulb is 200 W

• Potential difference (V) of the electric bulb is 240 V

⠀⠀⠀⠀

⠀⠀⠀⠀

To find:

⠀⠀⠀⠀

⠀⠀⠀⠀

Resistance (R) of the filament of bulb

⠀⠀⠀⠀

⠀⠀⠀⠀

Solution:

⠀⠀⠀⠀

We know that,

⠀⠀⠀⠀

P = V$^{2}$ /R

⠀⠀⠀⠀

$⇒ \: 200 = \frac{240 \times 240}{200} \\ \\ ⇒ R = \frac{240 \times 240}{200 } \\ \\ = 288 \: Ω$

⠀⠀⠀⠀

Thus, resistance of the filament of bulb is 288Ω.

Answer 2

Answer:

R =  166.66 Ω

Explanation:

Given,

P = 200 W

V = 240 V

To find : R = ?

V = I R

I = V / R ------> eq 1.

P = V I

  = V ( V / R )

P = V^2 / R

R = V^2 / P

  = 200^2 / 240

  = ( 200 x 200 ) / 240

R =  166.66 Ω

Therefore, the resistance of the filament of the bulb​ is 166.66 Ω.